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barxatty [35]
2 years ago
10

Koch traded Machine 1 for Machine 2 when the fair market value of both machines was $60,000. Koch originally purchased Machine 1

for $76,900, and Machine 1's adjusted basis was $40,950 at the time of the exchange. Machine 2's seller purchased it for $64,050 and Machine 2's adjusted basis was $55,950 at the time of the exchange. What is Koch's adjusted basis in machine 2 after the exchange?
Engineering
1 answer:
Mariana [72]2 years ago
3 0

Answer:

Koch's adjusted basis in machine 2 after the exchange is $60,000

Explanation:

given data

fair market value = $60,000

originally purchased Machine 1 = $76,900

Machine 1 adjusted basis = $40,950

Machine 2 seller purchase = $64,050

Machine 2 adjusted basis = $55,950

solution

As he exchanged machine for another at $60,000

and this exchanged in fair market

so adjusted basis =  $50,000

Adjusted basis is the price of the item that affects the factors that are considered price. These factors usually include taxes, depreciation value, and other costs of acquiring and maintaining a given item. Adjusted basis is important so the right amount to sell

Adjusted basis increases when a person deducts expenses from factor taxes and operating statements

so Koch's adjusted basis in machine 2 after the exchange is $60,000

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What is the reason for the development of new construction of materials for human use? (Select all that apply.)
salantis [7]

Answer:

All 4 could be justified.

Explanation:

They all represent ultimate improvement.

7 0
2 years ago
If the bolt head and the supporting bracket are made of the same material having a failure shear stress of 'Tra;i = 120 MPa, det
Nina [5.8K]

Answer:

P=361.91 KN

Explanation:

given data:

brackets and head of the screw are made of material with T_fail=120 Mpa

safety factor is F.S=2.5

maximum value of force P=??

<em>solution:</em>

to find the shear stress

                            T_allow=T_fail/F.S

                                         =120 Mpa/2.5

                                         =48 Mpa

we know that,

                               V=P

<u>Area for shear head:</u>

                              A(head)=π×d×t

                                           =π×0.04×0.075

                                           =0.003×πm^2

<u>Area for plate:</u>

                               A(plate)=π×d×t  

                                            =π×0.08×0.03

                                            =0.0024×πm^2

now we have to find shear stress for both head and plate

<u>For head:</u>

                                   T_allow=V/A(head)

                                    48 Mpa=P/0.003×π                 ..(V=P)

                                             P =48 Mpa×0.003×π

                                                =452.16 KN

<u>For plate:</u>

                                   T_allow=V/A(plate)

                                    48 Mpa=P/0.0024×π                 ..(V=P)

                                             P =48 Mpa×0.0024×π

                                                =361.91 KN

the boundary load is obtained as the minimum value of force P for all three cases. so the solution is

                                                P=361.91 KN

note:

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7 0
3 years ago
A silicon carbide plate fractured in bending when a blunt load was applied to the plate center. The distance between the fractur
Amanda [17]

Question in order:

A silicon carbide plate fractures in bending when a blunt load was applied to the plate center. The distance between the fracture origin and the mirror/mist boundary on the fracture surface was 0.796 mm. To determine the stress used to break the plate, three samples of the same material were tested and produced the following. What is the estimate of the stress present at the time of fracture for the original plate?

Mirror Radius (mm) Bending Failure Stress (MPa)

0.603                         225

0.203                         368

0.162                         442

Answer:

191 MPa

Explanation:

Failure stress of bending is Inversely proportional to the mirror radius

Bending Stress = \frac{1}{(Mirror Radius)^{n}}

At mirror radius 1 = 0.603 mm   Bending stress 1 = 225 Mpa..............(1)

At mirror radius 2 = 0.203 mm  Bending stress 2 = 368 Mpa...............(2)

At mirror radius 3 = 0.162 mm   Bending stress 3 = 442 Mpa...............(3)

comparing case 1 and 2 using the above equation

\frac{Stress 1}{Stress 2} = ({\frac{Radius 2}{Radius 1}})^{n_1}

\frac{225}{368} = ({\frac{0.203}{0.603}})^{n_1}

0.6114 = (0.3366)^{n_1}

Taking the natural logarithm of both side

ln(0.6114) = n ln(0.3366)

n₁ = ln(0.6114)/ln(0.3366)

n₁ = 0.452

comparing case 2 and 3 using the above equation

\frac{Stress 2}{Stress 3} = ({\frac{Radius 3}{Radius 2}})^{n_2}

\frac{368}{442} = ({\frac{0.162}{0.203}})^{n_2}

0.8326 = (0.7980)^{n_2}

Taking the natural logarithm of both side

ln(0.8326) = n ln(0.7980)

n₂ = ln(0.8326)/ln(0.7980)

n₂ = 0.821

comparing case 1 and 3 using the above equation

\frac{Stress 1}{Stress 3} = ({\frac{Radius 3}{Radius 1}})^{n_3}

\frac{225}{442} = ({\frac{0.162}{0.603}})^{n_3}

0.5090 = (0.2687)^{n_3}

Taking the natural logarithm of both side

ln(0.5090) = n ln(0.2687)

n₃ = ln(0.5090)/ln(0.2687)

n₃ = 0.514

average for n

n = \frac{n_1 + n_2 + n_3}{3}

n = \frac{0.452 +0.821 + 0.514}{3}

n = 0.596

Hence to get bending stress x at mirror radius 0.796

\frac{Stress x}{Stress 3} = ({\frac{Radius 3}{Radius x}})^{0.596}

\frac{Stress x}{225} = ({\frac{0.603}{0.796}})^{0.596}

\frac{Stress x}{225} = 0.8475

stress x = 191 MPa

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